Solve for $x$ and $y$ using elimination. $\begin{align*}-8x+3y &= 1 \\ 3x-2y &= 1\end{align*}$
Explanation: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $2$ and the bottom equation by $3$ $\begin{align*}-16x+6y &= 2\\ 9x-6y &= 3\end{align*}$ Add the top and bottom equations. $-7x = 5$ Divide both sides by $-7$ and reduce as necessary. $x = -\dfrac{5}{7}$ Substitute $-\dfrac{5}{7}$ for $x$ in the top equation. $-8( -\dfrac{5}{7})+3y = 1$ $\dfrac{40}{7}+3y = 1$ $3y = -\dfrac{33}{7}$ $y = -\dfrac{11}{7}$ The solution is $\enspace x = -\dfrac{5}{7}, \enspace y = -\dfrac{11}{7}$.